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anonymous
 5 years ago
please help! find the absolute max and min of f(x)=x^312x on [0,4]
anonymous
 5 years ago
please help! find the absolute max and min of f(x)=x^312x on [0,4]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y(dx/dy) = 3x^2 12 getting the critical numbers: 3x^2  12 = 0 3x^2 = 12 x^2 = 12/3 x^2 = 4 x = +2 substitute it to the original value y1=(2)^312(2) y1=16 y2=(2)^312(2) y2=16 get the y value of the interval y3 = (0)^3  12(0) y3 =0 y4 = (4)^3  12(4) y4 = 16 biggest is 16 now we know that abs max is 16 at x = 4 and x = 2 smalles is 16 so abs min is 16 at x=2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0find the derivative of f(x) and solve it when it's equal to 0 f'(x)=3x^212 \[3x^212=0 \implies 3x^2=12 \implies x^2=4\] x=2 or x=2 but 2 is not in [0,4] all you have to do now is to check the values of the function at x=2, and at the border points x=0, x=4 the maximum among them is the maximum in the interval, and same is applied to the minimum

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yea i forgot x= 2 cant be considered~~ but nevertheless 16 is still the abs maximum ^^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(2)=2^312(2)=824=16 f(0)=0 f(4)=4^312(4)=16 so the absolute maximum is 16 at x=4 and the absolute minimum is 16 at x=2
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